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no-shadow

Disallow variable declarations from shadowing variables declared in the outer scope.

This rule extends the base eslint/no-shadow rule. It adds support for TypeScript's this parameters and global augmentation, and adds options for TypeScript features.

How to Use

.eslintrc.cjs
module.exports = {
"rules": {
// Note: you must disable the base rule as it can report incorrect errors
"no-shadow": "off",
"@typescript-eslint/no-shadow": "error"
}
};

Try this rule in the playground ↗

Options

See eslint/no-shadow's options.

This rule adds the following options:

interface Options extends BaseNoShadowOptions {
ignoreTypeValueShadow?: boolean;
ignoreFunctionTypeParameterNameValueShadow?: boolean;
}

const defaultOptions: Options = {
...baseNoShadowDefaultOptions,
ignoreTypeValueShadow: true,
ignoreFunctionTypeParameterNameValueShadow: true,
};

ignoreTypeValueShadow

Whether to ignore types named the same as a variable. This is generally safe because you cannot use variables in type locations without a typeof operator, so there's little risk of confusion.

Examples of correct code with { ignoreTypeValueShadow: true }:

type Foo = number;
interface Bar {
prop: number;
}

function f() {
const Foo = 1;
const Bar = 'test';
}
Open in Playground
note

Shadowing specifically refers to two identical identifiers that are in different, nested scopes. This is different from redeclaration, which is when two identical identifiers are in the same scope. Redeclaration is covered by the no-redeclare rule instead.

ignoreFunctionTypeParameterNameValueShadow

Whether to ignore function parameters named the same as a variable.

Each of a function type's arguments creates a value variable within the scope of the function type. This is done so that you can reference the type later using the typeof operator:

type Func = (test: string) => typeof test;

declare const fn: Func;
const result = fn('str'); // typeof result === string

This means that function type arguments shadow value variable names in parent scopes:

let test = 1;
type TestType = typeof test; // === number
type Func = (test: string) => typeof test; // this "test" references the argument, not the variable

declare const fn: Func;
const result = fn('str'); // typeof result === string

If you do not use the typeof operator in a function type return type position, you can safely turn this option on.

Examples of correct code with { ignoreFunctionTypeParameterNameValueShadow: true }:

const test = 1;
type Func = (test: string) => typeof test;
Open in Playground

FAQ

Why does the rule report on enum members that share the same name as a variable in a parent scope?

Reporting on this case isn't a bug - it is completely intentional and correct reporting! The rule reports due to a relatively unknown feature of enums - enum members create a variable within the enum scope so that they can be referenced within the enum without a qualifier.

To illustrate this with an example:

const A = 2;
enum Test {
A = 1,
B = A,
}

console.log(Test.B);
// what should be logged?

Naively looking at the above code, it might look like the log should output 2, because the outer variable A's value is 2 - however, the code instead outputs 1, which is the value of Test.A. This is because the unqualified code B = A is equivalent to the fully-qualified code B = Test.A. Due to this behavior, the enum member has shadowed the outer variable declaration.

Resources

Taken with ❤️ from ESLint core.